Problem: Determine how many solutions exist for the system of equations. ${6x+2y = 10}$ ${3x+y = -8}$
Convert both equations to slope-intercept form: ${6x+2y = 10}$ $6x{-6x} + 2y = 10{-6x}$ $2y = 10-6x$ $y = 5-3x$ ${y = -3x+5}$ ${3x+y = -8}$ $3x{-3x} + y = -8{-3x}$ $y = -8-3x$ ${y = -3x-8}$ Just by looking at both equations in slope-intercept form, what can you determine? ${y = -3x+5}$ ${y = -3x-8}$ Both equations have the same slope with different y-intercepts. This means the equations are parallel. ${1}$ ${2}$ ${3}$ ${4}$ ${5}$ ${6}$ ${7}$ ${8}$ ${9}$ ${\llap{-}2}$ ${\llap{-}3}$ ${\llap{-}4}$ ${\llap{-}5}$ ${\llap{-}6}$ ${\llap{-}7}$ ${\llap{-}8}$ ${\llap{-}9}$ ${1}$ ${2}$ ${3}$ ${4}$ ${5}$ ${6}$ ${7}$ ${8}$ ${9}$ ${\llap{-}2}$ ${\llap{-}3}$ ${\llap{-}4}$ ${\llap{-}5}$ ${\llap{-}6}$ ${\llap{-}7}$ ${\llap{-}8}$ ${\llap{-}9}$ Parallel lines never intersect, thus there are NO SOLUTIONS.